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When tossed, a certain coin has equal probability of landing on either side. If the coin is tossed 3 times, what is the probability that it will land on the same side each time?
Let's break down what this problem is asking us in plain English. We have a fair coin, which means it has an equal chance of landing on heads or tails - just like flipping a quarter. We're going to toss this coin 3 times in a row.
The question asks: what's the probability that all 3 tosses will land on the same side? This means we want either:
- All 3 tosses to be heads, OR
- All 3 tosses to be tails
We don't care which side it is, just that all 3 tosses match each other.
Process Skill: TRANSLATE - Converting the problem language into clear mathematical understanding
When we toss a coin 3 times, let's think about what could happen. Each toss can be either heads (H) or tails (T). Let's list out every possible sequence:
First toss H: HHH, HHT, HTH, HTT
First toss T: THH, THT, TTH, TTT
So our complete list of equally likely outcomes is:
HHH, HHT, HTH, HTT, THH, THT, TTH, TTT
That gives us 8 total possible outcomes. This makes sense because each toss has 2 possibilities, and we have 3 tosses: \(2 \times 2 \times 2 = 8\) total outcomes.
Now let's look at our list and find the outcomes where all 3 tosses land on the same side:
From our list: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT
The favorable outcomes are:
- HHH (all heads)
- TTT (all tails)
So we have exactly 2 favorable outcomes out of our 8 total possible outcomes.
Process Skill: CONSIDER ALL CASES - Systematically identifying all scenarios that meet our criteria
Probability is simply the ratio of favorable outcomes to total possible outcomes.
In everyday terms: out of the 8 equally likely ways the coin tosses could turn out, exactly 2 of them give us what we want (all same side).
Probability = Favorable outcomes ÷ Total outcomes
Probability = \(2 ÷ 8 = \frac{1}{4}\)
The probability that the coin will land on the same side each time when tossed 3 times is \(\frac{1}{4}\).
Looking at our answer choices, this matches choice B: \(\frac{1}{4}\).
We can verify this makes sense: getting all heads or all tails should be relatively unlikely (less than \(\frac{1}{2}\)), but not extremely rare, and \(\frac{1}{4}\) (or 25%) seems reasonable for this scenario.
Students often misread this as "the probability of getting heads all three times" or "the probability of getting tails all three times," focusing on only one specific outcome rather than understanding that we want EITHER all heads OR all tails. This leads them to calculate only \(\frac{1}{8}\) instead of considering both favorable scenarios.
Some students think that after getting heads on the first toss, the probability changes for subsequent tosses to get "the same side." They might incorrectly set up the problem as \(\mathrm{P(H)} \times \mathrm{P(H|first\ was\ H)} \times \mathrm{P(H|first\ two\ were\ H)}\), not recognizing that each coin toss is independent with probability \(\frac{1}{2}\).
When listing all possible outcomes, students frequently miss some combinations or double-count others. For example, they might write down only 6 outcomes instead of all 8, or incorrectly list outcomes like "HHT, HTH, THH" thinking these are the only ways to get two heads and one tail.
Even when students correctly identify 2 favorable outcomes out of 8 total, they may make simple fraction errors. Common mistakes include writing \(\frac{2}{8} = \frac{1}{3}\) (incorrect simplification) or calculating \(\frac{2}{8}\) without reducing to \(\frac{1}{4}\).
Students who calculated \(\mathrm{P(HHH)} = \frac{1}{8}\) correctly but forgot to add \(\mathrm{P(TTT)} = \frac{1}{8}\) often select choice A (\(\frac{1}{8}\)) instead of the correct answer B (\(\frac{1}{4}\)). They solved half the problem correctly but failed to combine both favorable cases in their final answer selection.